We have $C$ sysmetric block matrices $W^c \in \mathbb{R}^{N_c\times N_c}$, where $c \in \lbrace 1,2,\cdots,C\rbrace$. We have the entire matrix consisting of $W^c$. We denote $W\in \mathbb{R}^{N\times N}$ as follows: $$ W=\begin{bmatrix} W^1 & \cdots & 0 \\ \vdots & \ddots & \vdots \\ 0 & \cdots & W^C \end{bmatrix} $$ Could anyone prove the following formulation: $$ \sum_{c=1}^{C}\sum_{i,j=1}^{N_c}\|\mathbf{x}_i^c-\mathbf{x}_j^c\|^2W_{ij}^c=\sum_{i,j=1}^{N}\|\mathbf{x}_i-\mathbf{x}_j\|^2W_{ij} $$ where $\mathbf{x}_i,\mathbf{x}_j\in \mathbb{R}^n$, $\mathcal{X}=\{\mathbf{x}_1^1,\mathbf{x}_2^1,\cdots,\mathbf{x}_{N_1}^1,\cdots,\mathbf{x}_1^C,\mathbf{x}_2^C,\cdots,\mathbf{x}_{N_C}^C\}$,$\mathbf{x}_i\in \mathcal{X},i\in \{1,2,\cdots,N\}$; where $N_1+N_2+\cdots+N_C=N$.
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2 It comes down to book-keeping. Here's one way to see it: $$ \sum_{i,j=1}^{N}\|\mathbf{x}_i-\mathbf{x}_j\|^2W_{ij} - \sum_{c=1}^{C}\sum_{i,j=1}^{N_c}\|\mathbf{x}_i^c-\mathbf{x}_j^c\|^2W_{ij}^c =\\ \sum_{1 \leq c\neq d \leq C} \sum_{i=1}^{N_c}\sum_{j=1}^{N_d} \|\mathbf{x}_i^c-\mathbf{x}_j^d\|^2W[N_1 + \cdots + N_c + i,N_1 + \cdots + N_d + j] $$ However, $W[N_1 + \cdots + N_c + i,N_1 + \cdots + N_d + j] = 0$
- $\begingroup$ Do you mean the following: $\mathbf{W}[N_1+\cdots+N_c+i,N_1+\cdots+N_d+i]\in \mathbf{W}$ and $\mathbf{W}[N_1+\cdots+N_c+i,N_1+\cdots+N_d+i]\ni \mathbf{W}^c$ $\endgroup$Dajiang Lei– Dajiang Lei2015-02-03 10:50:45 +00:00Commented Feb 3, 2015 at 10:50
- $\begingroup$ I mean that the entry of $W$ in the entry $(N_1 + \cdots + N_c + i,N_1 + \cdots + N_d + j)$ is $0$ because this entry is not in any of the matrices that form the block diagonal. $\endgroup$Ben Grossmann– Ben Grossmann2015-02-03 11:51:39 +00:00Commented Feb 3, 2015 at 11:51